For the equation x^4+x^3-4x^2+x-1=0,the ratio | vedclass

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(C) Given equation: $x^4+x^3-4x^2+x-1=0$. Factoring the polynomial: $x^4-1+x^3+x-4x^2 = (x^2-1)(x^2+1)+x(x^2+1)-4x^2 = (x^2+1)(x^2-1+x)-4x^2 = (x^2+1)

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$9: 1$

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(C) Given equation: $x^4+x^3-4x^2+x-1=0$. Factoring the polynomial: $x^4-1+x^3+x-4x^2 = (x^2-1)(x^2+1)+x(x^2+1)-4x^2 = (x^2+1)(x^2-1+x)-4x^2 = (x^2+1)(x^2+x-1)-4x^2$. Alternatively,notice $x=1$ is a root: $(x-1)(x^3+2x^2-2x+1)=0$. Further factoring: $(x-1)(x+1)(x^2+x-1)=0$. The roots are $x_1=1, x_2=-1, x_3=\frac{-1+\sqrt{5}}{2}, x_4=\frac{-1-\sqrt{5}}{2}$. Sum of squares of roots: $\sum x_i^2 = (1)^2+(-1)^2+(\frac{-1+\sqrt{5}}{2})^2+(\frac{-1-\sqrt{5}}{2})^2 = 1+1+\frac{1-2\sqrt{5}+5}{4}+\frac{1+2\sqrt{5}+5}{4} = 2+\frac{6}{4}+\frac{6}{4} = 2+3 = 5$. Distinct roots are $1, -1, \frac{-1+\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}$. Product of distinct roots: $1 	imes (-1) 	imes (\frac{-1+\sqrt{5}}{2} 	imes \frac{-1-\sqrt{5}}{2}) = -1 	imes (\frac{1-5}{4}) = -1 	imes (-1) = 1$. Ratio: $5:1$. Wait,re-evaluating the polynomial $x^4+x^3-4x^2+x-1=0$. Sum of squares $\sum x_i^2 = (\sum x_i)^2 - 2\sum x_ix_j = (-1)^2 - 2(-4) = 1+8 = 9$. Product of distinct roots: $1 	imes (-1) 	imes (-1) = 1$. Ratio is $9:1$.

For the equation $x^4+x^3-4x^2+x-1=0$,the ratio of the sum of the squares of all the roots to the product of the distinct roots is

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